Forum Home > GMAT > Quantitative > Problem Solving (PS)
Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Dec 1410:00 AM EST
-12:00 PM EST
Think a 100% GMAT Verbal score is out of your reach? Target Test Prep will make you think again! Our course uses techniques such as topical study and spaced repetition to maximize knowledge retention and make studying simple and fun.
Dec 1411:00 AM IST
-01:00 PM IST
Attend this session to learn how to Deconstruct arguments, Pre-Think Author’s Assumptions, and apply Negation Test.
Dec 1310:00 AM EST
-12:00 PM EST
Have you ever wondered how to score a PERFECT 805 on the GMAT? Meet Julia, a banking professional who used the Target Test Prep course to achieve this incredible feat. Julia's story is nothing short of an inspiration.- Dec 15
11:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease. 
Dec 1608:00 AM PST
-11:59 PM PST
GMAT Club 12 Days of Christmas is a 4th Annual GMAT Club Winter Competition based on solving questions. This is the Winter GMAT competition on GMAT Club with an amazing opportunity to win over $40,000 worth of prizes!- Dec 17
09:00 AM PST
-10:00 AM PST
Join Manhattan Prep instructor Whitney Garner for a fun—and thorough—review of logic-based (non-math) problems, with a particular emphasis on Data Sufficiency and Two-Parts. - Dec 18
10:00 AM PST
-11:00 AM PST
Here is the essential guide to securing scholarships as an MBA student! In this video, we explore the various types of scholarships available, including need-based and merit-based options.
23 May 2022, 13:52
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
60% (02:03) correct
40%
(02:20)
wrong
based on 62
sessions
History
Date
Time
Result
Not Attempted Yet
$5000 is invested in one account that pays x percent simple interest per year, and $6000 is invested in a different account that pays y percent simple interest per year. If the amount of money in each account is equal after one year, what is the value of x?
(1) x − y = 22
(2) If $6000 had been invested at x percent simple interest, and $5000 had been invested at y percent simple interest, after one year one account would have contained $2420 more than the other
(1) x − y = 22
(2) If $6000 had been invested at x percent simple interest, and $5000 had been invested at y percent simple interest, after one year one account would have contained $2420 more than the other
23 May 2022, 15:27
Kudos
Bookmarks
nislam
$5000 is invested in one account that pays \(x\) percent simple interest per year, and $6000 is invested in a different account that pays \(y\) percent simple interest per year. If the amount of money in each account is equal after one year, what is the value of \(x\)?
1) \(x\) − \(y\) = 22
2) If $6000 had been invested at \(x\) percent simple interest, and $5000 had been
invested at \(y\) percent simple interest, after one year one account would have contained $2420 more than the other
1) \(x\) − \(y\) = 22
2) If $6000 had been invested at \(x\) percent simple interest, and $5000 had been
invested at \(y\) percent simple interest, after one year one account would have contained $2420 more than the other
Translating the question stem gives us 5000(1+x)=6000(1+y) and asks for the value of x. We can't cancel out y from the given equation, so we need to derive a second (non-equivalent equation) so we can set up simultaneous equations and solve for x. Note that we converted from x% and y% to both in decimal form. Just in case we need it later, let's simplify the given equation as follows.
5000+5000x=6000+6000y
5+5x=6+6y
5x=6y+1
x=1.2y+0.2
Evaluating statement (1) alone:
x-y=0.22 (again converting both to decimal form)
That's a second equation that is non-equivalent to the first. We COULD solve the simultaneous equations, but we don't need to for Data Sufficiency!
Does statement (1) alone give us enough information to find the value of x? Yes. AD
Evaluating statement (2) alone:
This one is a little tricky. The statement provides us two options:
A) 6000(1+x)=5000(1+y)+2420
B) 6000(1+x)+2420=5000(1+y)
Both of those may LOOK like they'd work as simultaneous equations with the equation that we already have from the question stem, but B would result in negative values for x and y and the question is about earning simple interest, so x and y must be positive. That means B doesn't apply and we are left with only the equation from the question stem and A. Those are non-equivalent simultaneous equations, so we can solve for both variables...but we don't need to actually do it.
Does statement (2) alone give us enough information to find the value of x? Yes. D
Answer choice D.
23 May 2022, 19:42
Kudos
Bookmarks
nislam
$5000 is invested in one account that pays \(x\) percent simple interest per year, and $6000 is invested in a different account that pays \(y\) percent simple interest per year. If the amount of money in each account is equal after one year, what is the value of \(x\)?
1) \(x\) − \(y\) = 22
2) If $6000 had been invested at \(x\) percent simple interest, and $5000 had been
invested at \(y\) percent simple interest, after one year one account would have contained $2420 more than the other
1) \(x\) − \(y\) = 22
2) If $6000 had been invested at \(x\) percent simple interest, and $5000 had been
invested at \(y\) percent simple interest, after one year one account would have contained $2420 more than the other
The $5000 account begins $1000 behind the $6000 account.
Implication:
For the two accounts to end with the same amount, the interest earned by the $5000 investment must be $1000 greater than the interest earned by the $6000 investment:
\(\frac{x}{100} * 5000 = \frac{y}{100} * 6000 + 1000\)
\(50x = 60y + 1000\)
\(5x = 6y + 100\)
Note:
The resulting equation implies that x > y.
Statement 1:
Since we have two variables (x and y) and two distinct linear equations (5x=6y+100 and x-y=22), we can solve for the two variables.
Thus, the value of x can be determined.
SUFFICIENT.
Statement 2:
Since the prompt requires that x>y, the percentage applied here to the $5000 investment (y) is less than the percentage applied to the $6000 investment (x).
Implication:
It is not possible for the $5000 account to surpass the $6000 account by $2420.
The resulting amount for the $6000 investment must thus be $2420 greater than the resulting amount for the $5000 investment.
Since the $6000 investment begins $1000 ahead of the $5000 investment, a difference of $2420 will be yielded if the interest earned by the $6000 investment is $1420 greater than the interest earned by the $5000 investment:
\(\frac{x}{100} * 6000 = \frac{y}{100} * 5000 + 1420\)
\(60x = 50y + 1420\)
\(6x = 5y + 142\)
Since we have two variables (x and y) and two distinct linear equations (5x=6y+100 and 6x=5y+142), we can solve for the two variables.
Thus, the value of x can be determined.
SUFFICIENT.
